7. A corporation randomly selects 150 salespeople and find that 66% who have nev

Question

7. A corporation randomly selects 150 salespeople and find that 66% who have never taken a self-improvement course would like such a course. The firm did a similar study 10 years ago in which 60% of a random sample of 160 salespeople wanted a self-improvement course. The groups are assumed to be independent random samples. Conduct the test whether a greater proportion of workers would currently like to attend a self-improvement course than in the past, at 0.05 significance level. a. Use the critical value approach and interpret the test result. b. Use the p-value approach and interpret the test result

Explanation / Answer

p1 = 0.66

p2 = 0.6

H0: p1 = p2

H1: p1 > p2

Pooled sample proportion = (p1 * n1 + p2 * n2)/(n1 + n2)

= (0.66 * 150 + 0.6 * 160)/(150 + 160) = 0.629

SE = sqrt (P * (1 - P) * (1/n1 + 1/n2))

= sqrt (0.629 * 0.371 * (1/150 + 1/160))

= 0.0549

The test statistic Z = (p1 - p2)/SE

= (0.66 - 0.6)/0.0549

= 1.093

A) At alpha = 0.05, the critical value is 1.96

As the critical value is greater than the test statistic value (1.96 > 1.093), SO the null hypothesis is not rejected.

P-value = P(Z > 1.093)

= 1 - P(Z < 1.093)

= 1 - 0.8621

= 0.1379

As the p-value is greater than the significance level (0.1379 > 0.05), so the null hypothesis is not rejected.

So there is not sufficient evidence to support that a greater proportion of workers would currently like to attend a self-improvement course than in the past, at 0.05 significance level.

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